The question is as follows:
The matrix $\textbf{M}$ is given by
$\textbf{M}=\begin{pmatrix}
a & 1 & 1\\
1 & b & 1\\
1 & 1 & c\\
\end{pmatrix}$, where $a, b$ and $c$ are the roots of the equation $t^3-5t^2-5t+10=0$.
$\,\textbf{(i)}$ Show that the eigenvalues of $\textbf{M}$ are all real.
$\textbf{(ii)}$ Show that if $\lambda$ is an eigenvalue of $\textbf{M}$ then $|\lambda|<6$.
My attempt at a solution:
Using the symmetric functions of the roots of the $t$-equation, I obtained
\begin{cases}
a+b+c=5,\\
ab+bc+ac=-5,\\
abc=-10
\end{cases}
From which I used these values to obtain the characteristic equation of $\textbf{M}$ to be $\lambda^3-5\lambda^2+8\lambda-13=0$. Denoting the eigenvalues of $\textbf{M}$ by $\lambda_i$ for $i=1, 2, 3$, I argued that since $\sum_{i=1}^{3}{\lambda_i^2}=5^2-2(8)=9>0\Rightarrow$ All $\lambda_i$ are real
How would one proceed to prove the statement in part $\textbf{(ii)}$? I used $\mathrm{tr}(\textbf{M})=\lambda_1+\lambda_2+\lambda_3$ and $\det(\textbf{M})=\lambda_1\lambda_2\lambda_3$ to obtain
\begin{cases}
\lambda_1+\lambda_2+\lambda_3=5\\
\lambda_1\lambda_2\lambda_3=-15
\end{cases}
though I struggle to see how the given statement can be proved.
[$\textbf{Note}$: This is an old A-level Further Mathematics question so I'd really appreciate it if the response can be simplified a bit to suit my level of knowledge]
Best Answer
Let $f(t)=(t-a)(t-b)(t-c)=t^3-5t^2-5t+10$. The characteristic polynomial of $M$ is therefore equal to \begin{aligned} p(t)&=f(t)-2-(t-a)-(t-b)-(t-c)\\ &=f(t)-3t+(a+b+c-2)\\ &=t^3-5t^2-8t+13. \end{aligned} Since $p(-3)=-35,\,p(0)=13,\,p(2)=-15$ and $p(6)=1$, $M$ has three real eigenvalues, one on $(-3,0)$, one on $(0,2)$ and the other on $(2,6)$. It follows that all of these eigenvalues have moduli $<6$.